How it works
Motors, transformers and other inductive loads draw reactive power as well as real power, so the supply must carry more current than the kW alone would need. A low power factor means bigger cables, more loss and, in many tariffs, a penalty.
Capacitors supply reactive power locally. The bank size is the difference between the reactive power before and after: Qc = kW × (tan φ1 − tan φ2). Raising the power factor lowers the kVA and the line current for the same kW.
The capacitance is given for a delta connected three phase bank, or for a single capacitor on a single phase supply. Large motor loads, harmonics and switching need a proper design, for example detuned reactors and an automatic controller.
Formulas used
- Qc (kVAr) = kW × (tan φ1 − tan φ2), with φ = arccos(power factor)
- kVA = kW ÷ power factor
- Current = kVA × 1000 ÷ (√3 × V) three phase, kVA × 1000 ÷ V single phase
- Capacitance per phase (delta) C = Qc × 1000 ÷ (3 × 2π f × V²)
Worked example
A 100 kW load at power factor 0.75 raised to 0.95 needs 100 × (0.882 − 0.329) = 55.3 kVAr. The kVA falls from 133.3 to 105.3 and the current on 415 V from 185.5 A to 146.4 A.
The calculator opens with these example values, so you can compare its answer with the text above.
Frequently asked questions
How do I calculate the capacitor size to improve power factor?
Multiply the kW by the difference of tan φ before and after, where φ is the angle whose cosine is the power factor.
What is a good target power factor?
Many utilities ask for 0.95 or better. Going to 1.0 is rarely worth the cost and can over-correct at light load.
Why does a low power factor cost money?
It raises the current for the same useful power, which needs larger equipment and increases losses. Many tariffs also charge for kVA or penalise a low power factor.
Can I use any capacitor?
No. Use capacitors rated for the voltage and duty, with protection, and have the bank designed for your load and harmonics.